answer:1. Let there be several lines on a plane, no two of which are parallel. 2. Assume that not all lines pass through a single point. 3. Consider the set ( M ) of points where at least two of the given lines intersect. 4. Choose one of the lines, say ( m ). 5. Select the intersection point ( P ) from set ( M ) that lies closest to the line ( m ) but is not on ( m ). Let’s consider two cases: Case 1: Exactly two lines intersect at ( P ) - If exactly two lines intersect at point ( P ), then ( P ) is a point where precisely two given lines, say ( l_1 ) and ( l_2 ), intersect. This completes the proof for this case. [ boxed{} ] Case 2: More than two lines intersect at ( P ) - Suppose three lines ( PA, PB, ) and ( PC ) intersect at point ( P ), where points ( A, B,) and ( C ) lie on the line ( m ). - Without loss of generality, let ( B ) lie between points ( A ) and ( C ) on the line ( m ). - If through point ( B ), exactly two lines ( l_1 ) and ( l_2 ) pass, the proof is concluded. [ boxed{} ] Otherwise, another line ( k ) (distinct from ( m ) and ( PB )) passes through point ( B ). - Line ( k ) must intersect one of the segments ( PA ) or ( PC ) of the triangle ( APC ) in some point ( K ). - This point ( K ) belongs to the set ( M ) since it is an intersection of at least two distinct lines. - The distance from ( K ) to the line ( m ) is less than the distance from ( P ) to the line ( m ). Thus, there exists a point ( K ) that contradicts the selection of ( P ) as the closest intersection point to line ( m ). Therefore, this contradiction proves that: [ boxed{text{Either exactly two lines intersect at some point, or all lines intersect at a single point.}} ]

answer:1. **Define the centers and points of tangency:** Let the centers of the circles mathcal{K}_1, mathcal{K}_2, mathcal{K}_3 be A, B, C respectively. Relabel the point of tangency between mathcal{K}_1 and mathcal{K}_2 as D. Define E and F as the points of tangency between mathcal{K}_1 and mathcal{K}_3, and mathcal{K}_2 and mathcal{K}_3 respectively. Let M be the intersection of the external common tangent of mathcal{K}_1 and mathcal{K}_2 with mathcal{K}_3. 2. **Concurrency of lines:** Due to the tangency and the properties of the circles, the lines CF, BE, AD concur at a point. Let R be the intersection of the internal common tangent of mathcal{K}_1 and mathcal{K}_2 with BC. Notice that R is the harmonic conjugate of D with respect to B and C. 3. **Harmonic division:** Since R and D are harmonic conjugates, we have the relation: [ RE cdot RF = RD^2 = RP cdot RQ ] This implies that the circle passing through E, F, D is the incircle of triangle ABC. 4. **Tangency condition:** The circle Gamma passing through P, Q, D is tangent to BC at R. This is because RE cdot RF = RP cdot RQ. 5. **Perpendicularity and angles:** Let G be the foot of the perpendicular from D to EF. Notice that: [ angle QT G = angle PDM ] due to the tangency and the fact that angle MDC = angle MDB = 90^circ. 6. **Diameter and midpoint:** Extend DM to hit Gamma at H. Since TH is a diameter of Gamma, we need to prove that M is the midpoint of segment TH. 7. **Inversion:** Invert the problem around D with an arbitrary radius. In the inverted diagram, mathcal{K}_3' and Gamma' are reflections with respect to the line P'Q'. This implies that H' is the midpoint of TM', completing the proof. [ blacksquare ]

answer:# Solution: Part (Ⅰ): Discussing the monotonicity of f(x) Given the function f(x)=frac{1}{2}ax^2+(a-1)x-ln x, which is defined on the domain (0,+infty). To discuss the monotonicity, we first find the derivative of f(x) with respect to x: [ f'(x) = frac{d}{dx}left(frac{1}{2}ax^2+(a-1)x-ln xright) = ax + a - 1 - frac{1}{x} = frac{(x+1)(ax-1)}{x}. ] - **Case 1:** If a leqslant 0, then for all x in (0,+infty), we have f'(x) < 0. This implies that f(x) is monotonically decreasing on (0,+infty). - **Case 2:** If a > 0, then: - For x in (0,frac{1}{a}), we have f'(x) < 0, indicating that f(x) is monotonically decreasing. - For x in (frac{1}{a},+infty), we have f'(x) > 0, indicating that f(x) is monotonically increasing. Therefore, summarizing the above cases: - If a leqslant 0, f(x) is monotonically decreasing on (0,+infty). - If a > 0, f(x) is monotonically decreasing on (0,frac{1}{a}) and monotonically increasing on (frac{1}{a},+infty). Part (Ⅱ): Proving f(x)geqslant 2-frac{3}{{2a}} when a>0 From Part (Ⅰ), when a > 0, f(x) reaches its minimum at x = frac{1}{a}. Thus, we evaluate f(x) at this point: [ fleft(frac{1}{a}right) = 1 - frac{1}{2a} - lnfrac{1}{a}. ] To prove f(x) geqslant 2 - frac{3}{2a}, we consider the inequality: [ 1 - frac{1}{2a} - lnfrac{1}{a} geqslant 2 - frac{3}{2a} implies frac{1}{a} - lnfrac{1}{a} - 1 geqslant 0. ] Let g(x) = x - ln x - 1 and find its derivative: [ g'(x) = 1 - frac{1}{x}. ] - For x in (0,1), g'(x) < 0, indicating g(x) is monotonically decreasing. - For x in (1,+infty), g'(x) > 0, indicating g(x) is monotonically increasing. At x = 1, g(x) reaches its minimum value, which is g(1) = 0. Therefore, for x > 0, we have g(x) geqslant 0. This implies that when a > 0, frac{1}{a} - lnfrac{1}{a} - 1 geqslant 0, i.e., f(x) geqslant 2 - frac{3}{2a}. Hence, we conclude that when a > 0, f(x) geqslant 2 - frac{3}{2a}, which can be encapsulated as: [ boxed{f(x) geqslant 2 - frac{3}{2a}}. ]

answer:Let alpha = arccos z, so 0 leq alpha leq pi and cos alpha = z. Now, [ sin alpha = sqrt{1 - z^2}, ] so, using the above: [ sin (arccos z) = sqrt{1-z^2}. ] Then, compute: [ tan(sin (arccos z)) = tan sqrt{1-z^2}. ] Therefore, the equation becomes: [ cos left( arctan left(tan sqrt{1-z^2}right) right) = z. ] Since arctan(tan x) = x for x within (-frac{pi}{2},frac{pi}{2}), [ cos (sqrt{1-z^2}) = z. ] Let z = sin beta for beta in [0, pi], we then find: [ cos(sqrt{1-sin^2 beta}) = sin beta. ] Since sqrt{1-sin^2 beta} = cos beta, and cos(cos beta) = sin beta, we directly simplify to: [ z = sin left(arccos left(frac{1}{sqrt{2}} right) right). ] As arccosleft(frac{1}{sqrt{2}}right) = frac{pi}{4}, [ z = sinleft(frac{pi}{4}right) = frac{sqrt{2}}{2}, ] we verify the condition 0 leq z leq 1 is satisfied. So there is exactly boxed{1} positive solution, namely boxed{z = frac{sqrt{2}}{2}}.